Theorem nss 3988

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1866	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		dfss2 3912	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	xchbinxr 335	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 223	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1540  ∃wex 1786   ∈ wcel 2110   ⊆ wss 3892

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1545  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-v 3433  df-in 3899  df-ss 3909

This theorem is referenced by:  grur1  10577  psslinpr  10788  reclem2pr  10805  mreexexlem2d  17352  prmcyg  19493  filconn  23032  alexsubALTlem4  23199  wilthlem2  26216  shne0i  29806  erdszelem10  33158  fundmpss  33736  ntrneineine1lem  41664  nssrex  42606  nssd  42625  nsssmfmbf  44282